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Leni [432]
3 years ago
13

Solve 8y(y−2)+8=0 by using the Quadratic Formula.

Mathematics
1 answer:
klio [65]3 years ago
8 0

Answer:

x=1

Step-by-step explanation:

First, we must simplify it to standard form:

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The volume of a cube is 27 cubic inches. Which expression represents s, the length of a side of the cube?
NeTakaya

volume of a cube = S^3

which would be S x S x S

so answer is D

8 0
3 years ago
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Complete the square to solve the equation below.
Vinil7 [7]

9514 1404 393

Answer:

  B.  x = -6, x = 4

Step-by-step explanation:

The constant on the left wants to be 1, the square of half the x-coefficient. We can get it to be that value by adding 8 to the equation.

  x² +2x +1 = 25

  (x +1)² = 25 . . . . . . . rewrite the left side as a square

  x +1 = ±√25 = ±5 . . . . take the square root

  x = -1 ±5 . . . . . . . . . subtract 1

  x = -6 or x = 4

3 0
3 years ago
What are these numbers in order from greatest to least
DerKrebs [107]
Answer : Square root of 21 , -4 2/3, -30/7 , -31/6 .

Explanation :

-4 2/3 = -10/3 = -3.3333
Square root of 21 = 4.58258
-31/6 = -5.1666
-30/7 = 4.2857

And arrange them from greatest to least and there we have the answer.

Hope this helps you understand !!!!
4 0
4 years ago
The dye dilution method is used to measure cardiac output with 3 mg of dye. The dye concentrations, in mg/L, are modeled by c(t)
Lemur [1.5K]

Answer:

Cardiac output:F=0.055 L\s

Step-by-step explanation:

Given : The dye dilution method is used to measure cardiac output with 3 mg of dye.

To Find : Find the cardiac output.

Solution:

Formula of cardiac output:F=\frac{A}{\int\limits^T_0 {c(t)} \, dt} ---1

A = 3 mg

\int\limits^T_0 {c(t)} \, dt =\int\limits^{10}_0 {20te^{-0.06t}} \, dt

Do, integration by parts

[\int{20te^{-0.6t}} \, dt]^{10}_0=[20t\int{e^{-0.6t} \,dt}-\int[\frac{d[20t]}{dt}\int {e^{-0.6t} \, dt]dt]^{10}_0

[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20}{0.6}\int {e^{-0.6t} \,dt]^{10}_0

[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20e^{-0.6t}}{(0.6)^2}]^{10}_{0}

[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-200e^{-6}}{0.6}+\frac{20e^{-6}}{(0.6)^2}]+\frac{20}{(0.60^2}

[\int{20te^{-0.6t}} \, dt]^{10}_0=\frac{20(1-e^{-6}}{(0.6)^2}-\frac{200e^{-6}}{0.6}

[\int{20te^{-0.6t}} \, dt]^{10}_0\sim {54.49}

Substitute the value in 1

Cardiac output:F=\frac{3}{54.49}

Cardiac output:F=0.055 L\s

Hence Cardiac output:F=0.055 L\s

4 0
3 years ago
the sum of two mixed numbers is 5 1/2. the differenceof mixed numbers is 3. both numbers have a denominator of 4. what are the t
Rudik [331]

Answer:

ok

Step-by-step explanation:

5'm

5 0
3 years ago
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